/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 A 2.0 L flask holds 0.40 g of he... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

Short Answer

Expert verified
(D) It will increase because the gas molecules will be more dispersed in the larger flask. As the volume increases while the temperature remains constant, the entropy increases due to greater dispersal of gas molecules.

Step by step solution

01

Understanding the concept of Entropy

Entropy is a measure of the disorder or randomness of a system. In this exercise, we are considering a case where a gas is transferred from a smaller to a larger container, keeping the temperature constant.
02

Impact of Volume on Entropy

If the volume in which the gas is contained increases, there will be more space for the gas particles to move about. This would lead to higher randomness or disorder in the gas distribution, which in turn will lead to an increase in entropy.
03

Identifying the right answer

Based on our analysis in the previous steps, the answer is (D) It will increase because the gas molecules will be more dispersed in the larger flask. This is because as volume increases, at a constant temperature, the gas particles have increased space to move leading to an increase in entropy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas Laws
Gas laws help us understand the behavior of gases in different scenarios. They describe how variables like volume, pressure, temperature, and quantity are interrelated in gases. Imagine you're studying how a balloon behaves when you take it outside on a chilly day or how it expands when it's heated.

One important gas law relevant here is Boyle’s Law, which states that the volume of gas is inversely proportional to pressure at a constant temperature. This means if you increase the volume available to a gas, its pressure will decrease, assuming the temperature remains unchanged.

Another crucial principle is the Ideal Gas Law, expressed as \( PV = nRT \), where \( P \) represents pressure, \( V \) represents volume, \( n \) is the amount of gas in moles, \( R \) is the ideal gas constant, and \( T \) is temperature.

In this exercise, since the helium gas is transferred to a larger container without changing the temperature, the volume increases – potentially altering the dynamics of pressure and molecular movement, which are all interconnected by these laws. Understanding gas laws allows us to predict how gases behave under changing conditions with accuracy.
Disorder and Randomness
Entropy is intricately linked to the concepts of disorder and randomness. It represents the degree of chaos within a system. Imagine you have a neatly stacked pile of cards. If you shuffle them, you've increased the disorder of the cards – in essence, you’ve increased their entropy.

When applying this understanding to gases, if a gas occupies a larger space, its molecules have more room to move around freely and in a random manner. This increased freedom signifies that the molecules are more "disordered" compared to when they were cramped in a smaller space.

This disorder is what increases the entropy of the system when the volume of the gas container is increased, as the molecules can now occupy a greater number of positions. Thus, transferring helium to a larger flask increases its entropy due to the greater possible configurations and randomness of the particles in the larger volume.
Temperature and Volume Relationship
While temperature remains constant in this scenario, it's important to note how temperature generally affects gas volume – a key part of the study of thermodynamics and gas behavior.

According to Charles’s Law, the volume of a gas is directly proportional to its temperature when the pressure is kept constant. This means if you increase the temperature of a gas, its volume also increases if the pressure doesn't change.

However, in the presented exercise, since the temperature remains constant, it does not directly influence the volume increase caused by evacuating the helium into the larger flask. This allows us to focus on the concept of expanding volume as a means to understand entropy, free from other changing variables like temperature.

Often, understanding the relationship between temperature and volume can shed light on how gases will respond in dynamic situations, helping reinforce concepts like why and how entropy increases with volume.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) If the reaction above took place at standard temperature and pressure and 150 grams of \(\mathrm{CaCO}_{3}(\mathrm{s})\) were consumed, what was the volume of \(\mathrm{CO}_{2}(g)\) produced at STP? (A) 11 L (B) 22 L (C) 34 L (D) 45 L

In an experiment 2 moles of \(\mathrm{H}_{2}(g)\) and 1 mole of \(\mathrm{O}_{2}(g)\) were completely reacted, according to the following equation in a sealed container of constant volume and temperature: $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$$ If the initial pressure in the container before the reaction is denoted as \(P_{i}\) which of the following expressions gives the final pressure, assuming ideal gas behavior? (A) \(P_{i}\) (B) 2\(P_{i}\) (C) \((3 / 2) P_{i}\) (D) \((2 / 3) P_{i}\)

A student added 1 liter of a 1.0\(M\) \(\mathrm{KCl}\) solution to 1 liter of a 1.0 \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution. A lead chloride precipitate formed, and nearly all of the lead ions disappeared from the solution. Which of the following lists the ions remaining in the solution in order of decreasing concentration? (A) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]\) (B) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{K}^{+}\right]\) (C) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{NO}_{3}^{-}\right]\) (D) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]\)

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. As \(\Delta T\) increases, what happens to the equilibrium constant and why? (A) The equilibrium constant increases because more products are created. (B) The equilibrium constant increases because the rate of the forward reaction increases. (C) The equilibrium constant decreases because the equilibrium shifts to the left. (D) The value for the equilibrium constant is unaffected by temperature and will not change.

Nitrogen’s electronegativity value is between those of phosphorus and oxygen. Which of the following correctly describes the relationship between the three values? (A) The value for nitrogen is less than that of phosphorus because nitrogen is larger, but greater than that of oxygen because nitrogen has a greater effective nuclear charge. (B) The value for nitrogen is less than that of phosphorus because nitrogen has fewer protons, but greater than that of oxygen because nitrogen has fewer valence electrons. (C) The value for nitrogen is greater than that of phosphorus because nitrogen has fewer electrons, but less than that of oxygen because nitrogen is smaller. (D) The value for nitrogen is greater than that of phosphorus because nitrogen is smaller, but less than that of oxygen because nitrogen has a smaller effective nuclear charge.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.